Proof with Powers of 2 and a ProductDate: 10/22/2008 at 17:59:08 From: Off Subject: Any power of 2 can be multiplied with result digits 1 2 only I need to prove that for any power of 2 there is a number that when you multiply the two you get a number with digits 1 and 2 only. For example: 4*3 = 12 8*14 = 112 64*33 = 2112 Thank you, Off Date: 10/24/2008 at 22:52:34 From: Doctor Vogler Subject: Re: Any power of 2 can be multiplied with result digits 1 2 only Hi, Thanks for writing to Dr. Math. That's an interesting question. I would start with the final digits, since if you have the n'th power of 2 (that is, 2^n), then the last n digits are determined. So I would argue that there is an infinite sequence a_n with a_1 = 2 a_2 = 12 a_3 = 112 a_4 = 2112 a_5 = 22112 a_6 = 122112 a_7 = 2122112 ... such that a_n is an n-digit number consisting only of 1's and 2's, such that a_n is divisible by 2^n. You can define the sequence recursively: if (a_n)/2^n is even, then a_{n+1} is just 2 prepended to (stuck on the left side of) a_n; if (a_n)/2^n is odd, then a_{n+1} is just 1 prepended to a_n. This is easy enough to prove: If (a_n)/2^n is even, then 2^(n+1) divides a_n, and 2^(n+1) divides 2*10^n, and therefore 2^(n+1) divides a_n + 2*10^n. Similarly, if (a_n)/2^n is odd, then (10^n)/2^n = 5^n is also odd, so (a_n + 10^n)/2^n is even, which means that a_n + 10^n is divisible by 2^(n+1). And that does it! If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 10/25/2008 at 20:44:39 From: Off Subject: Thank you (Any power of 2 can be multiplied with result digits 1 2 only) Thank you Dr. Vogler for your help! I have spent about a month and could not crack it and you did it in a day. Amazing... |
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